Answer
$x\approx 2.7944$
Work Step by Step
Recall the definition of a logarithm (pg. 451):
$\log_{b}{x}=y$ iff $b^y=x$
Applying this definition to our equation, we get:
$\log_{12}{20}=4-x$
Next, recall the change of base formula (pg. 464):
$\log_{b}{m}=\frac{\log_{c}{m}}{\log_{c}{b}}$
Applying this property, we get:
$\frac{\log_{10}{20}}{\log_{10}{12}}=4-x$
$x=4-\frac{\log_{10}{20}}{\log_{10}{12}}$
$x\approx 2.7944$
We confirm that the answer works:
$12^{4-2.7944}=20$
$12^{1.2056}=20$
$20=20$
